# Fixed points of $\phi_a(z)$

Prove that $\phi_a(z)=\frac{a-z}{1-\bar az}$ , $0<|a|<1$ has exactly two fixed points ; one inside the unit disc and the other outside the unit disc.

Putting $\phi_a(z)=z$ I find that there are exactly two fixed points which are $$\frac{1\pm \sqrt{1-|a|^2}}{\bar a}.$$

I am unable to find out the second part.

Let $z_1=\frac{1- \sqrt{1-|a|^2}}{\bar a}$. From $0<|a|<1$ , we get $z_1<\frac{1}{\bar a}=\frac{a}{|a|^2}$. Then , $|z_1|<\frac{1}{|a|}$. But how I show that $|z_1|<1$ so that I can say that $z_1$ lies inside the circle ?

You are solving the quadratic $$\def\abar{\overline a}\abar z^2-2z+a=0\ .$$ By the usual product-of-roots formula we have $$z_1z_2=a\,/\,\abar$$ and so $$|z_1||z_2|=|a\,/\,\abar|=1\ .$$ Therefore one of $z_1$ and $z_2$ has modulus less than $1$, the other greater.

(It could be that both have modulus equal to $1$, but it is not very hard to see that this is impossible.)

• You beat me to the punch line! +1 Jul 20 '15 at 3:26

$z_1\overline{a}=1-\sqrt{1-x^2}, x = |a|\to (|z_1||\overline{a}|)^2 =1-2\sqrt{1-x^2}+1-x^2\to bx^2= 2-x^2-2\sqrt{1-x^2}$. Thus we can prove: $2-x^2-2\sqrt{1-x^2} < x^2 \iff 1-x^2 < \sqrt{1-x^2}$ which is true since $0 < 1-x^2 < 1$. Thus $b = |z_1| < 1$